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padilas [110]
3 years ago
14

Triangle MNP is transformed according to the rule T(4,–1). What is the y-coordinate of M”?

Mathematics
2 answers:
weqwewe [10]3 years ago
8 0
-1 because x is on the left and y is always on the right
Nitella [24]3 years ago
8 0

Answer:

ITS 3!!!!!! Just took the text

Step-by-step explanation: 3!!!!!!

You might be interested in
Given that ABCD is a rhombus, what is the value of x?<br>(3x - 26)​
ExtremeBDS [4]
<h2>The required "option B) 20.75\°" is correct.</h2>

Step-by-step explanation:

You can refer figure:

brainly.com/question/4386376

To find, the value of x = ?

We know that,

Consecutive angles are supplementary

The diagonals bisect the angles.

2x\°+2(3x+7)\°=180\°

Divided by 2 both sides, we get

⇒ x\°+(3x+7)\°=90\°

⇒ x\°+3x+7\°=90\°

⇒ 4x=90\°-7\°

⇒ 4x = 83\°

⇒ x=\dfrac{83\°}{4}

⇒ x=20.75\°

∴ The value of x =20.75\°

Thus, the required "option B) 20.75\°" is correct.

7 0
3 years ago
Read 2 more answers
Find all solutions to the following quadratic equations, and write each equation in factored form.
dexar [7]

Answer:

(a) The solutions are: x=5i,\:x=-5i

(b) The solutions are: x=3i,\:x=-3i

(c) The solutions are: x=i-2,\:x=-i-2

(d) The solutions are: x=-\frac{3}{2}+i\frac{\sqrt{7}}{2},\:x=-\frac{3}{2}-i\frac{\sqrt{7}}{2}

(e) The solutions are: x=1,\:x=-1,\:x=\sqrt{5}i,\:x=-\sqrt{5}i

(f) The solutions are: x=1

(g) The solutions are: x=0,\:x=1,\:x=-2

(h) The solutions are: x=2,\:x=2i,\:x=-2i

Step-by-step explanation:

To find the solutions of these quadratic equations you must:

(a) For x^2+25=0

\mathrm{Subtract\:}25\mathrm{\:from\:both\:sides}\\x^2+25-25=0-25

\mathrm{Simplify}\\x^2=-25

\mathrm{For\:}x^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}x=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}\\\\x=\sqrt{-25},\:x=-\sqrt{-25}

\mathrm{Simplify}\:\sqrt{-25}\\\\\mathrm{Apply\:radical\:rule}:\quad \sqrt{-a}=\sqrt{-1}\sqrt{a}\\\\\sqrt{-25}=\sqrt{-1}\sqrt{25}\\\\\mathrm{Apply\:imaginary\:number\:rule}:\quad \sqrt{-1}=i\\\\\sqrt{-25}=\sqrt{25}i\\\\\sqrt{-25}=5i

-\sqrt{-25}=-5i

The solutions are: x=5i,\:x=-5i

(b) For -x^2-16=-7

-x^2-16+16=-7+16\\-x^2=9\\\frac{-x^2}{-1}=\frac{9}{-1}\\x^2=-9\\\\\mathrm{For\:}x^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}x=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}\\x=\sqrt{-9},\:x=-\sqrt{-9}

The solutions are: x=3i,\:x=-3i

(c) For \left(x+2\right)^2+1=0

\left(x+2\right)^2+1-1=0-1\\\left(x+2\right)^2=-1\\\mathrm{For\:}\left(g\left(x\right)\right)^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}g\left(x\right)=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}\\\\x+2=\sqrt{-1}\\x+2=i\\x=i-2\\\\x+2=-\sqrt{-1}\\x+2=-i\\x=-i-2

The solutions are: x=i-2,\:x=-i-2

(d) For \left(x+2\right)^2=x

\mathrm{Expand\:}\left(x+2\right)^2= x^2+4x+4

x^2+4x+4=x\\x^2+4x+4-x=x-x\\x^2+3x+4=0

For a quadratic equation of the form ax^2+bx+c=0 the solutions are:

x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}

\mathrm{For\:}\quad a=1,\:b=3,\:c=4:\quad x_{1,\:2}=\frac{-3\pm \sqrt{3^2-4\cdot \:1\cdot \:4}}{2\cdot \:1}

x_1=\frac{-3+\sqrt{3^2-4\cdot \:1\cdot \:4}}{2\cdot \:1}=\quad -\frac{3}{2}+i\frac{\sqrt{7}}{2}\\\\x_2=\frac{-3-\sqrt{3^2-4\cdot \:1\cdot \:4}}{2\cdot \:1}=\quad -\frac{3}{2}-i\frac{\sqrt{7}}{2}

The solutions are: x=-\frac{3}{2}+i\frac{\sqrt{7}}{2},\:x=-\frac{3}{2}-i\frac{\sqrt{7}}{2}

(e) For \left(x^2+1\right)^2+2\left(x^2+1\right)-8=0

\left(x^2+1\right)^2= x^4+2x^2+1\\\\2\left(x^2+1\right)= 2x^2+2\\\\x^4+2x^2+1+2x^2+2-8\\x^4+4x^2-5

\mathrm{Rewrite\:the\:equation\:with\:}u=x^2\mathrm{\:and\:}u^2=x^4\\u^2+4u-5=0\\\\\mathrm{Solve\:with\:the\:quadratic\:equation}\:u^2+4u-5=0

u_1=\frac{-4+\sqrt{4^2-4\cdot \:1\left(-5\right)}}{2\cdot \:1}=\quad 1\\\\u_2=\frac{-4-\sqrt{4^2-4\cdot \:1\left(-5\right)}}{2\cdot \:1}=\quad -5

\mathrm{Substitute\:back}\:u=x^2,\:\mathrm{solve\:for}\:x\\\\\mathrm{Solve\:}\:x^2=1=\quad x=1,\:x=-1\\\\\mathrm{Solve\:}\:x^2=-5=\quad x=\sqrt{5}i,\:x=-\sqrt{5}i

The solutions are: x=1,\:x=-1,\:x=\sqrt{5}i,\:x=-\sqrt{5}i

(f) For \left(2x-1\right)^2=\left(x+1\right)^2-3

\left(2x-1\right)^2=\quad 4x^2-4x+1\\\left(x+1\right)^2-3=\quad x^2+2x-2\\\\4x^2-4x+1=x^2+2x-2\\4x^2-4x+1+2=x^2+2x-2+2\\4x^2-4x+3=x^2+2x\\4x^2-4x+3-2x=x^2+2x-2x\\4x^2-6x+3=x^2\\4x^2-6x+3-x^2=x^2-x^2\\3x^2-6x+3=0

\mathrm{For\:}\quad a=3,\:b=-6,\:c=3:\quad x_{1,\:2}=\frac{-\left(-6\right)\pm \sqrt{\left(-6\right)^2-4\cdot \:3\cdot \:3}}{2\cdot \:3}\\\\x_{1,\:2}=\frac{-\left(-6\right)\pm \sqrt{0}}{2\cdot \:3}\\x=\frac{-\left(-6\right)}{2\cdot \:3}\\x=1

The solutions are: x=1

(g) For x^3+x^2-2x=0

x^3+x^2-2x=x\left(x^2+x-2\right)\\\\x^2+x-2:\quad \left(x-1\right)\left(x+2\right)\\\\x^3+x^2-2x=x\left(x-1\right)\left(x+2\right)=0

Using the Zero Factor Theorem: = 0 if and only if = 0 or = 0

x=0\\x-1=0:\quad x=1\\x+2=0:\quad x=-2

The solutions are: x=0,\:x=1,\:x=-2

(h) For x^3-2x^2+4x-8=0

x^3-2x^2+4x-8=\left(x^3-2x^2\right)+\left(4x-8\right)\\x^3-2x^2+4x-8=x^2\left(x-2\right)+4\left(x-2\right)\\x^3-2x^2+4x-8=\left(x-2\right)\left(x^2+4\right)

Using the Zero Factor Theorem: = 0 if and only if = 0 or = 0

x-2=0:\quad x=2\\x^2+4=0:\quad x=2i,\:x=-2i

The solutions are: x=2,\:x=2i,\:x=-2i

3 0
3 years ago
24 x 12 1/2 x 12 3/4 what is the volume of the fish tank
hjlf

Answer:

The volume is 3825 cubic units

Step-by-step explanation:

To find the volume of the fish tank multiply out the dimensions.

We were given the dimensions as:

24 \times 12 \frac{1}{2}  \times 12 \frac{3}{4}

Change the numbers to improper fractions

24 \times  \frac{25}{2}  \times \frac{51}{4}

We cancel out to get

3 \times 25 \times 51

3825

5 0
3 years ago
The sum of two numbers is equal to 63, and their difference is equal to 12. Find the numbers ​
Ainat [17]

Answer:

25.5   and   37.5

Step-by-step explanation:

x+y=63

x-y = 12    then   x = 12 + y      sub this into the first equation

(12+y)    + y = 63

12 + 2y = 63

2y = 51

y = 25.5     then x = 37.5

6 0
2 years ago
PLEASE HELP I GIVE THANKS
dezoksy [38]
For cylinders to be similar, the ratio between their height and radius must be equal.Our cylinder's ratio is :
\frac { length }{ radius } =\frac { 15 }{ 3 } \\ =5

The only cylinder with same ratio is the one at the choice B.

\frac { length }{ radius } \frac { 25 }{ 5 } =5 
6 0
3 years ago
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